Question Number 12261 by frank ntulah last updated on 17/Apr/17
$${find}\:{the}\:{value}\:{of}\:{a}\:{b}\:{and}\:{c} \\ $$$$\:{a}+{b}+{c}=\mathrm{4} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{66} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{280} \\ $$
Answered by mrW1 last updated on 17/Apr/17
$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ca}=\mathrm{4}^{\mathrm{2}} =\mathrm{16} \\ $$$$\Rightarrow{ab}+{bc}+{ca}=\frac{\mathrm{16}−\mathrm{66}}{\mathrm{2}}=−\mathrm{25} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\mathrm{6}{abc}+\mathrm{3}\left({a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {a}+{b}^{\mathrm{2}} {c}+{c}^{\mathrm{2}} {a}+{c}^{\mathrm{2}} {b}\right) \\ $$$$={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\mathrm{3}\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)−\mathrm{3}{abc} \\ $$$$=\mathrm{280}+\mathrm{3}×\mathrm{4}×\left(−\mathrm{25}\right)−\mathrm{3}{abc}=\mathrm{4}^{\mathrm{3}} =\mathrm{64} \\ $$$$\Rightarrow{abc}=\frac{\mathrm{280}+\mathrm{3}×\mathrm{4}×\left(−\mathrm{25}\right)−\mathrm{64}}{\mathrm{3}}=−\mathrm{28} \\ $$$$ \\ $$$${a},{b},{c}\:{are}\:{the}\:{roots}\:{of}\:{following}\:{cubic}\:{eqn}. \\ $$$${x}^{\mathrm{3}} −\left({a}+{b}+{c}\right){x}^{\mathrm{2}} +\left({ab}+{bc}+{ca}\right){x}−{abc}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{25}{x}+\mathrm{28}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{1}\right)\left({x}+\mathrm{4}\right)\left({x}−\mathrm{7}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =−\mathrm{4},\:{x}_{\mathrm{2}} =\mathrm{1},\:{x}_{\mathrm{3}} =\mathrm{7} \\ $$$${possible}\:{solutions}\:{for}\:{a},{b},{c}\:{are}: \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}=\begin{pmatrix}{−\mathrm{4}}\\{\mathrm{1}}\\{\mathrm{7}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}=\begin{pmatrix}{−\mathrm{4}}\\{\mathrm{7}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{4}}\\{\mathrm{7}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{7}}\\{−\mathrm{4}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}=\begin{pmatrix}{\mathrm{7}}\\{−\mathrm{4}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}=\begin{pmatrix}{\mathrm{7}}\\{\mathrm{1}}\\{−\mathrm{4}}\end{pmatrix} \\ $$
Commented by frank ntulah last updated on 17/Apr/17
$${thank}\:{you}\:{sir}\:{you}\:{make}\:{me}\:{happy} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by tawa last updated on 17/Apr/17
$$\mathrm{Yes}\:\mathrm{sir},\:\mathrm{check}\:\mathrm{line}\:\mathrm{4}\:……\:\:\:\:\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{c}^{\mathrm{3}} \:….. \\ $$
Commented by mrW1 last updated on 17/Apr/17
$${Thanks}.\:{You}\:{are}\:{right}.\:{I}\:{have}\:{corrected}. \\ $$
Commented by tawa last updated on 17/Apr/17
$$\mathrm{Yes}\:\mathrm{sir}.\:\mathrm{Thank}\:\mathrm{you}.\:\mathrm{i}\:\mathrm{always}\:\mathrm{follow}\:\mathrm{your}\:\mathrm{steps}\:\mathrm{very}\:\mathrm{well}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$